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User blog:Cloudy176/Simulating PEGG: part 2
It's been 100 days since Psi's Ever-Growing Googolism (PEGG) was defined, and that was where I left off in the previous simulating PEGG blog post. Today, I decided to post what will happen after that day, which I actually made a few days after writing the previous blog post but procrastinated to post it until now. (I'm writing this blog post to provide an excuse to play with Letter Notation expressions.) In this blog, instead of listing the values day by day, we'll skip days by a multiple of 5 to make the list shorter. For now, I'll list what happens when the letter transition occurs; I'll add the list of values between transitions later. This blog uses the modified definition of PEGG whose definition can be found on the previous blog post. You can verify the calculation with the list of 10000 random integers produced by RANDOM.ORG, or compare these values with the actual values of PEGG. Some links: Definition of PEGG, Definition of Letter Notation: E~H J~N P, Definition of lexicographic ordering (LOS) The age of E to G In the previous blog post, we looked at the first 100 days of the simulated PEGG. Here's a recap: This is how it grows after the 100th day: The age of H In day 111 (2017-08-27), we switch from G to H. The new value of X is calculated as follows: G10.0311 = GE1.00134856... = GF1.00058528... = GG1.00025411... = H2.00011034... Therefore the new value of X is 2.00011. The values are updated to Y = "H", Z = 0.00343 and K = 1000000, and the simulation continues: The age of J In day 167 (2017-10-22), we switch from H to J. The new value of X is calculated as follows: H10.0709 = HE1.003068284... = HF1.001330499... = HG1.000577444... = HH1.000250708... = 5|2.000108868... = 5|2*5^0.000033821... = 4.000033821...|10 = J4.000033821... Therefore the new value of X is 4.000033. The values are updated to Y = "J", Z = 0.002401 and K = 10^7, and the simulation continues: The age of K In day 232 (2017-12-26), we switch from J to K. The new value of X is calculated as follows: J10.122583 = JE1.0052913464... = JF1.0022919442... = JG1.0009942398... = JJ1.0009942398... = K2.0004315783... Therefore the new value of X is 2.0004315. The values are updated to Y = "K", Z = 0.0016807 and K = 10^8, and the simulation continues (with 10 days per line instead of 5): The age of L In day 363 (2018-05-06), we switch from K to L. The new value of X is calculated as follows: K10.0946827 = KE1.00409267258... = KF1.00177379780... = KG1.00076966818... = KJ1.00076966818... = KK1.00033413407... = L2.00014508835... Therefore the new value of X is 2.00014508. The values are updated to Y = "L", Z = 0.00117649 and K = 10^9, and the simulation continues: The age of M In day 537 (2018-10-27), we switch from L to M. The new value of X is calculated as follows: L10.00498304 = LE1.000216356776... = LF1.000093952391... = LG1.000040801088... = LJ1.000040801088... = LK1.000017719326... = LL1.000007695337... = (1,3)|2.000003342030... = (1,3)|2*5^0.000001038259... = (1,2.000001038259...)|10 = M2.000001038259... Therefore the new value of X is 2.000001038. The values are updated to Y = "M", Z = 0.000823543 and K = 10^10, and the simulation continues: The age of N In day 812 (2019-07-29), we switch from M to N. The new value of X is calculated as follows: M10.056722207 = ME1.0024564539433... = MF1.0010655162320... = MG1.0004625014622... = MJ1.0004625014622... = MK1.0002008153978... = ML1.0000872042635... = (2,0)|(1,3)|1.0000378706792... = (2,0)|(2,0)|1.0000378706792... = (2,1)|2.0000164467156... = (2,0.0000051094385...)|10 = N2.0000005109438... Therefore the new value of X is 2.0000005109. The values are updated to Y = "N", Z = 0.0005764801 and K = 10^11, and the simulation continues: The age of P In day 1166 (2020-07-17), we switch from N to P. The new value of X is calculated as follows: N10.0165327815 = NE1.00071741669620... = NF1.00031145840301... = NG1.00013524360548... = NJ1.00013524360548... = NK1.00005873158013... = NL1.00002550605217... = (1,0,0)|(1,3)|1.00001107699645... = (1,0,0)|(2,0)|1.00001107699645... = (1,0,0)|(1,0,0)|1.00000481065179... = (1,0,1)|2.00000208923450... = (1,0,0.00000064905685...)|10 = (1.00000000649057...,0,0)|10 = P2.00000000281882... Therefore the new value of X is 2.00000000281. The values are updated to Y = "P", Z = 0.00040353607 and K = 10^12, and the simulation continues: What happens next? At this point we should switch from P to Q, however Q isn't defined yet for all real input. (section to be written) Category:Blog posts